EECE 310 Exam 2, Fall 2007

EECE 310 Electronics I – Section 1 (N. Sabah) Quiz 2 – December 8, 2007 One Hour and a Half – Closed Book 5:1 Penalty on Questions 1 to 7 ALL QUESTIONS MUST BE ANSWERED ON THE QUESTION SHEET NAME: ___________________________________________________ ID: ___________________________________________________ 7% 1. Given two pn junctions i) pn junction 1 having ND1 = 1016/cm3 and NA1 = 1014/cm3, and ii) pn junction 2 having ND2 = 1014/cm3 and NA2 = 1016/cm3. Which pn junction will have the larger magnitude of junction potential? A. pn junction 1 B. pn junction 2 C. Both pn junctions will have the same magnitude of junction potential D. None of the above Solution: The magnitude of the chemical potential difference is . Since this is the same in both cases, the magnitude of the electrical potential difference will be the same. 7% 2. NA and ND on the two sides of a pn junction are varied but their product is kept constant at 1032/cm6. For what NA and ND is the width of the depletion region W, a minimum? A. NA = ND B. NA > ND C. NA < ND D. NA = ND but W is a maximum E. None of the above 1/6 Solution: , where everything is constant except (NA + ND). The sum is a minimum when NA = ND. Thus, if y = NA + , where K = NAND, , which gives NA = ND. That this is a minimum is clear from . 7% 3. If the temperature increases, the open circuit voltage of a photocell: A. increases B. decreases C. stays the same Solution: – ISS, where, ISS = AJ Gx(Wd + Lh + Le). Wen iD = 0, Voc = . ISS changes weakly with T but IS increases exponentially with T. Hence, VOC decreases with T. 7% 4. A normal pn junction diode having  = 1 is paralleled with a Schottky diode of the same junction area. Which diode will carry the larger current? A. The normal pn junction diode. B. The Schottky diode. Solution: The Schottky diode equation is , and the equation for a normal diode having  = 1 is . IB0 >> IS, so that for the same vB, the Schottky diode will carry the larger current. 2/6 7% 5. In the circuit shown, VDD = 5 V, Vt = 1 V, C is initially discharged, and vI suddenly increases from zero to a steady value VDD. The final value of vC is: A. 5 V B. 4 V C. 3 V D. 1 V E. None of the above Solution: When vI increases from zero to VDD, with vC = 0 initially, vGS = VDD, iD flows, C charges, vC increases, and vGS and iD decrease. iD becomes zero when vGS = Vt. The final value of vC is therefore, VDD – Vt = 5 – 1 = 4V. 7% 6. The E-NMOS transistor has A / V2, , Vtn = 1 V and  = 0. If it is replaced by a D-NMOS transistor whose transfer characteristic is that of the E-NMOS transistor shifted by 3 V, the D-NMOS transistor will be: A. in the triode region B. in the saturation region C. cut-off D. None of the above Solution: VGS = 2 V, VtnD = 1 – 3 = -2 V; assuming transistor is in saturation, mA. This makes vDS < 0, which means the assumption is wrong and the transistor is in the triode region. To confirm this, iD and vDS can be found by solving the equations: and vDS + iD = 5. This gives mA, vDS = 1.13 V; vDS < vGS – VtnD, which confirms that transistor is in triode region. 3/6 1 k +5 V 3 k 2 k VDD vI vC + – + – C 7% 7. The E-NMOS transistor of the preceding problem is replaced by a matched E-PMOS transistor with its drain grounded and its source connected to the 1 k resistor. The EPMOS transistor will be: A. in the triode region B. in the saturation region C. cut-off D. None of the above Solution: VSG = 3 – iD; VSD = 5 – iD. Assuming that the transistor is in the saturation region, . This gives iD = = 0.764 mA and 5.24 mA. The latter is discarded because it makes vSG and vSD negative. Hence, iD = 0.764 mA; vSG = 2.24 V and vSD = 4.24 V, which confirms that the transistor is in saturation. 17% 8. In the circuit shown, VSS = -5 V and R = 2 k. The transistor has A / V2, , Vtn = 1 V and  = 0. 12% (a) Determine the transistor current. 5% (b) If and the transistor has the same operating conditions, determine the incremental output resistance of the transistor as seen by R. Solution: (a) VGS + 2ID – 5 = 0, or VGS = 5 – 2ID. The transistor is in saturation, because its drain and gate are at the same voltage. . Substituting for VGS, , or , which gives ID = 1 mA or 4 mA. If ID = 4 mA, VGS = -3 V, which is impossible. Hence, ID = 1 mA and VGS = 3 V. (b) mA / V. this makes ro = 100 k. gm = 0.521.03 = 1.03 mS, Output resistance is rO||(1/gm) = 0.97 k. 4/6 R VSS R VSS ID 1 k +5 V 3 k 2 k 17% 9. In the circuit shown, Q1 and Q2 have mA / V2, Vt = 0.8 V, , 0.05 V-1, (W/L)n1 = 10, and (W/L)n2 = 20. Determine IO. Solution: Q1 is in saturation, so 1 = ; hence, V. Assuming Q2 is in saturation, IO = = 2 + 0.1VDS2. From KVL, VDS2 + 0.5IO = 5, or VDS2 = 5 – 0.5IO. Substituting, IO = 2.38 mA and VDS2 = 3.81 V, which confirms that Q2 is in saturation. 17% 10. The transistors of a CMOS inverter operating from a VDD = 3 V supply have the following parameters: E-NMOS: 120 A/V2, (W/L)n = 4 Vtn = 0.6 V  = 0 E-PMOS: 50 A/V2, (W/L)p = 6 Vtp = -0.7 V  = 0 Determine the input voltage vI that makes the inverter operate on the vertical segment of its transfer characteristic. Solution: Over the vertical segment of the transfer characteristic, both transistors operate in the saturation region, with equal currents. For Qn: iDn = . For Qp: . Equating the currents: , or . This gives V, or -5.81 V. The negative value is discarded, so that vI = 1.35 V. 5/6 500  +5V Q1 Q2 1 mA IO Useful Equations – ISS, where, ISS = AJ Gx(Wd + Lh + Le) For the Schottky diode: (saturation region) (triode region) 6/6